Let $f$ be a complex function $f(z) = \frac{z^2 +1}{z(z-i)^3}$. It needs to be put jnto Laurent's series, so that it will converge in a set $D = \{ g(z) \mid Re(z) > 0 \}$, where $g(z) = \frac{2iz}{z-1}$.
I've been struggling with finding this set $D$. So the singularities for $f$ are $z = 0$ and $z = i$. Where should I even have the center of the series?
One option is to write $g$ as a composition of “simple” transformations: $$ g(z) = 2i \left( 1 + \frac{1}{z-1}\right) $$ and then successively determine the image of the right half-plane.
Another option is to use that fact that $g$ is a Möbius transformation: It maps the imaginary axis to a circle (or line), and that image is uniquely determined by the image of three distinct points. From $$ g(-i) = -1+i, g(0) = 0, g(+i) = 1+i $$ you can conclude that the imaginary axis is mapped to the circle with center $i$ and radius $1$. It follows that the right half-plane is mapped to the inside or outside of that circle. Since $g(1) = \infty$, we have $$ D = \{ g(z) : \operatorname{Re}(z) > 0 \} = \{ w : |w-i| > 1 \} \, . $$
That is an annulus with center $i$, inner radius $1$ and outer radius $\infty$. Your task is now to determine the Laurent series of $f$ in that annulus.